# How do you find the equation of the tangent line and normal line to the curve y = (5+6x)^2 at the point (-4,361)?

Apr 2, 2015

1) Equation of Tangent

Step1:

The tangent is a line so it has the form of a line:
$y - {y}_{0} = m \left(x - {x}_{0}\right)$
$\left({x}_{0} , {y}_{0}\right)$ represents the point at which the tangent touches the curve, in this case we have $\left(- 4 , 361\right)$
Also the gradient, $m = \frac{\mathrm{dy}}{\mathrm{dx}}$
So,
$y = {\left(5 + 6 x\right)}^{2}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cdot \left(5 + 6 x\right) \times 6$
$= 12 \left(5 + 6 x\right)$

Step2:

At the point $\left(- 4 , 361\right)$ we substitute the value of $x = - 4$ and this becomes,

$\frac{\mathrm{dy}}{\mathrm{dx}} = 12 \left(5 + 6 \cdot \left(- 4\right)\right)$
$= 12 \cdot \left(- 19\right) = - 228$

So, $m = - 228$

Step3:

Putting the point $\left(- 4 , 361\right)$ and gradient $m = - 228$ in the slope-point form:

$\implies y - 361 = - 288 \left(x - \left(- 4\right)\right)$
hence equation of tangent is $y - 1 = - 288 \left(x + 4\right)$

Equation of normal

The equation of normal is also a line of the same form as the tangent.

The only difference is at the level of the gradient.

We can deduce the gradient of the normal from the fact that the normal is a line perpendicular to the tangent

This implies the product of their gradients is $- 1$

So let ${m}_{N}$ be gradient of normal and ${m}_{T}$ be gradient of tangent.

$\implies {m}_{N} \times {m}_{T} = - 1$
$\implies {m}_{N} = - \frac{1}{{m}_{T}} = - \frac{1}{- 288} = \frac{1}{288}$

Hence equation of normal is $y - 361 = \left(\frac{1}{288}\right) \times \left(x + 4\right)$